Statistics — Grouped Data 📊

Mean ke teen methods, Median aur Mode — complete derivations aur worked examples.

📋 Grouped Data Kya Hai?

🏫 School Exam Example:

Ek school mein 200 students ke marks 0 se 100 tak hain. Agar hum har student ka mark alag se list karein — 200 numbers — bahut mushkil ho jaayega patterns dekhna. Isliye hum class intervals banate hain: 0-10, 10-20, 20-30... aur count karte hain kitne students har range mein hain. Yahi grouped frequency distribution hai!

Important Terms:
Class Interval: Range jaise 10-20 (lower = 10, upper = 20)
Class Width (h): Upper − Lower = 20−10 = 10
Class Mark (xᵢ): Middle value = (10+20)/2 = 15
Frequency (fᵢ): Us class mein kitne data points hain
Cumulative Frequency (cf): Current tak ki sari frequencies ka sum

Sample Data: Students ke Marks (50 students)

Class IntervalFrequency (fᵢ)Class Mark (xᵢ)fᵢxᵢCumulative f
10-25217.5352
25-40332.597.55
40-55747.5332.512
55-70662.537518
70-85677.546524
85-100692.555530
TotalΣf=30Σfx=1860

📊 Mean — 3 Methods

Method 1: Direct Method
Mean = Σfᵢxᵢ / Σfᵢ
✅ Best when numbers are small and easy to calculate
Direct Method on above data:
Mean = 1860 / 30 = 62
Method 2: Assumed Mean Method (Short Cut)
Mean = a + (Σfᵢdᵢ / Σfᵢ)
dᵢ = xᵢ − a  (a = assumed mean, usually middle class mark)
✅ Best when xᵢ values are large but differences dᵢ are small
Assumed Mean Method: a = 62.5 (middle class mark)
Classfᵢxᵢdᵢ=xᵢ−62.5fᵢdᵢ
10-25217.5−45−90
25-40332.5−30−90
40-55747.5−15−105
55-70662.500
70-85677.5+15+90
85-100692.5+30+180
Total30Σfᵢdᵢ = −15
Mean = 62.5 + (−15/30) = 62.5 − 0.5 = 62 ✓
Method 3: Step Deviation Method
Mean = a + h × (Σfᵢuᵢ / Σfᵢ)
uᵢ = (xᵢ − a) / h  (divide by class width h)
✅ Best when class width h is constant — simplest calculations
Step Deviation: a=62.5, h=15
uᵢ values: −3, −2, −1, 0, +1, +2
Σfᵢuᵢ = 2(−3)+3(−2)+7(−1)+6(0)+6(1)+6(2) = −6−6−7+0+6+12 = −1
Mean = 62.5 + 15 × (−1/30) = 62.5 − 0.5 = 62 ✓

📐 Median of Grouped Data

Median: Data ko 2 equal halves mein divide karne wala value.
Grouped data mein hum formula use karte hain kyunki exact values pata nahi hain.
Median Formula
Median = l + [ (n/2 − cf) / f ] × h

l = lower boundary of median class
n = Σf = total frequency
cf = cumulative frequency BEFORE median class
f = frequency OF median class
h = class width
Median class kaise dhundein?
1. Cumulative frequency table banao
2. n/2 calculate karo (n = total frequency)
3. Jis class mein cumulative frequency n/2 ko cross kare — woh median class
Example: Upar wali table se Median nikalo (n=30, n/2=15)
Cumulative frequencies: 2, 5, 12, 18, 24, 30
n/2 = 15. Kahan cross hota hai? Class 55-70 mein (cumulative = 18 ≥ 15, pehle 12 hai)
Median class: 55-70
l=55, n/2=15, cf=12, f=6, h=15
Median = 55 + [(15−12)/6] × 15 = 55 + (3/6)×15 = 55 + 7.5 = 62.5

📈 Mode of Grouped Data

Mode Formula
Mode = l + [ (f₁−f₀) / (2f₁−f₀−f₂) ] × h

l = lower boundary of modal class (class with highest frequency)
f₁ = frequency of modal class
f₀ = frequency of class BEFORE modal class
f₂ = frequency of class AFTER modal class
h = class width
Example: Upar wali table se Mode nikalo
Modal class = 40-55 (f=7, highest frequency)
l=40, f₁=7, f₀=3 (class 25-40), f₂=6 (class 55-70), h=15
Mode = 40 + [(7−3)/(14−3−6)] × 15 = 40 + (4/5) × 15 = 40 + 12 = 52

🔗 Empirical Relation

Mode = 3 × Median − 2 × Mean
Verify: Mode = 3(62.5) − 2(62) = 187.5 − 124 = 63.5 ≈ 52 (ye approximately correct hoga, exact nahi kyunki yeh empirical formula hai)

📊 Ogive (Cumulative Frequency Graph)

Ogive = cumulative frequency ka graph. Do types:
Less than Ogive: Upper class boundary vs cumulative frequency (badh ta jaata hai)
More than Ogive: Lower class boundary vs (total−cumulative) frequency (ghat ta jaata hai)
Dono Ogive ka intersection = Median!
cf x Less than More than Median!

📝 Practice Questions

Q1. Neeche diya hua data hai. Direct method se mean nikalo.

Wages (₹)Workers
100-12010
120-14015
140-16020
160-18022
180-20018
200-22015
xᵢ: 110,130,150,170,190,210. fᵢxᵢ: 1100,1950,3000,3740,3420,3150. Σfᵢxᵢ=16360. Σf=100.
Mean = 163.6

Q2. Upar wali table se modal class kaunsi hai? Mode nikalo.

Highest f=22 → Modal class = 160-180. l=160, f₁=22, f₀=20, f₂=18, h=20.
Mode = 160 + [(22−20)/(44−20−18)]×20 = 160 + (2/6)×20 = 160+6.67 = 166.67

Q3. Ek hospital mein 100 patients ka data: Q2 ki table se Median nikalo.

n=100, n/2=50. Cumulative frequencies: 10,25,45,67,85,100. n/2=50, crosses at class 160-180 (cf before=45, f=22).
Median = 160 + [(50−45)/22]×20 = 160 + (5/22)×20 = 160+4.55 = 164.55

Q4. Agar mean = 62 aur mode = 52 ho, to median kya hoga? (Empirical relation use karo)

Mode = 3×Median − 2×Mean → 52 = 3×Median − 124 → 3×Median = 176 → Median = 58.67
🎮 Stats Lab → Next: Probability →